Question

Differentiate sin ^-1( 2ax√1- a^2x^2 with respect to cos^-1{√1-a^2x^2}

Answer 1

Step-by-step explanation:

Given that:

Differentiate sin ^-1( 2ax√1- a^2x^2 with respect to cos^-1{√1-a^2x^2}

Solution:

To differentiate

${sin}^{ - 1} (2ax \sqrt{1 - {a}^{2} {x}^{2} } ) \\ w.r.t. \\ {cos}^{ - 1} ( \sqrt{1 - {a}^{2} {x}^{2} } )$

Let

$\bold{u = {sin}^{ - 1} (2ax \sqrt{1 - {a}^{2} {x}^{2} } ) }\\ and \\\bold{v = {cos}^{ - 1} ( \sqrt{1 - {a}^{2} {x}^{2} } )}$

now find du/dx and dv/dx

$let \\ \: sin \alpha = ax \\ so \\ cos \: \alpha = \sqrt{1 - {a}^{2} {x}^{2} }$

put these values in u and v

$u = {sin}^{ - 1} (2 \sin \alpha \cos\alpha ) \\ \\ sin2 \alpha = 2 \sin \alpha \cos\alpha \\ \\ u = {sin}^{ - 1} (sin \: 2 \alpha ) \\ \\ u = 2 \alpha$

put the value of alpha from assumption in terms of sin inverse

$u = 2 {sin}^{ - 1} (ax) \\ \\ differentiate \: u \: wrt \: x \\ \\ \bold{ \frac{du}{dx} = \frac{2a}{ \sqrt{1 - {a}^{2} {x}^{2} } } } \: \: \: \: \: ....eq1$

By the same way Differentiation v with respect to x,after substitution and simplification

$v = {cos}^{ - 1} ( \sqrt{1 - {sin}^{2} \alpha } ) \\ \\ v = {cos}^{ - 1} (cos \alpha ) \\ \\ v = \alpha \\ \\ v = {sin}^{ - 1} (ax) \\ \\ differentiate \: v \: wrt \: to \: x \\ \\ \bold{\frac{dv}{dx} = \frac{a}{ \sqrt{1 - {a}^{2} {x}^{2} } }} \: \: \: \: ..eq2$

Now divide eq1 by eq2

$\frac{du}{dv} = \frac{du}{dx} \times \frac{dx}{dv} \\ \\ = \frac{2a}{ \sqrt{1 - {a}^{2} {x}^{2} } } \times \frac{ \sqrt{1 - {a}^{2} {x}^{2} } }{a} \\ \\ \bold{\frac{du}{dv} = 2}$

Hope it helps you.

Answer 2

Answer:

du/dv=2

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